intuition for similar matrix

Sure. $P$ is a linear transformation which takes things to a new basis. after applying $B$, you bring your (transformed) basis back to where it used to be (transformed). If this is the same thing as your original transformation ($A$), then $A$ and $B$ must fundamentally be the same transform, in the different basis. I visualize it with a rotation example: rotate to a new basis ($P$), do your rotation $B$, and rotate back $(P^{-1})$. If this is the same as one rotation $A$, then they were the same.

$Start\rightarrow^P Rotated$

$\downarrow A\hspace{40 pt}\downarrow B$

$End \leftarrow^{P^{-1}} Twice$

If you end at the same place then $A$ and $B$ must be the same.

I always understood that the matrix $P$ is a change of basis matrix from the domain to itself. So as the codomain and the domain are the same vector space, the definition is saying $A$ and $B$ are similar if transforming by $A$ is the same as changing basis, then transforming via $B$, then changing back again. Hope it helps.

My current understanding, not sure this helps:

$B=PAP^{-1}$ means $BP=PA$, so if $A$ is a diagonal, $BP=PA$ is just the definition of eigenvetors, which is easy to understand. Because we know a matrix and its diagonal form are the same transformation under different basis, so to extrapolate to non-diagonal cases, we can define any square matrix of the form $B'P=PA'$ or $B'=PA'P^{-1}$ representing the same transformation under different basis, called similar matrix.

Edit: $B=PAP^{-1}$ means $A=P^{-1}BP$, and think of $P$ as the change of basis matrix, then $P^{-1}BP$ means change of basis first, then do the linear transformation, then change the basis back, so $A$ and $B$ represent the same linear transformation under different bases. See here: https://youtu.be/P2LTAUO1TdA?t=9m12s